calcuate the maximum numbers of moles and grams of H2S that can form when 158g of aluminum sulfide reacts with 131g of water:Al2S3 + H2O -gt;Al(OH)3 + H2S eq is unbalancedwhat mass of the excess reactant remains?
Well, the balanced equation is Al2S3 + 6 H2O 2 Al(OH)3 + 3 H2S mmAl2S3: molar mass 150.16 grams/mol mmH2O: molar mass 18.02 grams/mol mAl2S3:moles of Al2S3 1.052 moles mH20: moles of H20 7.27 moles The equation says that 3 moles of H2S form from every mole of Al2S3 and 6 moles of H2OIt looks like Al2S3 limits the reaction (we would need less than 6.3 moles of H2O for water to limit the reaction)So you will produce mH2S: moles H2S 31.052 3.15 mol H2S mmH2S: molar mass H2S 34.08 grams mass of H2S 107.4 grams This reaction will consume all the aluminum sulfideWe will consume 6.312 moles of H2O, leaving 17.3 grams (0.97 moles) of H2O.
put it in a bath, that should keep the current flowingdo you need it to work as well? i would probably just get a new one to be safe.
